Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{-r - 4}{-r^2 + 4r + 32} \times \dfrac{-5r^2 + 25r + 120}{r - 3} $
Solution: First factor out any common factors. $x = \dfrac{-(r + 4)}{-(r^2 - 4r - 32)} \times \dfrac{-5(r^2 - 5r - 24)}{r - 3} $ Then factor the quadratic expressions. $x = \dfrac {-(r + 4)} {-(r - 8)(r + 4)} \times \dfrac {-5(r - 8)(r + 3)} {r - 3} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {-(r + 4) \times -5(r - 8)(r + 3) } { -(r - 8)(r + 4) \times (r - 3)} $ $x = \dfrac {5(r - 8)(r + 3)(r + 4)} {-(r - 8)(r + 4)(r - 3)} $ Notice that $(r - 8)$ and $(r + 4)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {5\cancel{(r - 8)}(r + 3)(r + 4)} {-\cancel{(r - 8)}(r + 4)(r - 3)} $ We are dividing by $r - 8$ , so $r - 8 \neq 0$ Therefore, $r \neq 8$ $x = \dfrac {5\cancel{(r - 8)}(r + 3)\cancel{(r + 4)}} {-\cancel{(r - 8)}\cancel{(r + 4)}(r - 3)} $ We are dividing by $r + 4$ , so $r + 4 \neq 0$ Therefore, $r \neq -4$ $x = \dfrac {5(r + 3)} {-(r - 3)} $ $ x = \dfrac{-5(r + 3)}{r - 3}; r \neq 8; r \neq -4 $